Question: What is the extraneous solution to these equations? $\dfrac{x^2 + x}{x - 10} = \dfrac{6x + 14}{x - 10}$
Explanation: Multiply both sides by $x - 10$ $ \dfrac{x^2 + x}{x - 10} (x - 10) = \dfrac{6x + 14}{x - 10} (x - 10)$ $ x^2 + x = 6x + 14$ Subtract $6x + 14$ from both sides: $ x^2 + x - (6x + 14) = 6x + 14 - (6x + 14)$ $ x^2 + x - 6x - 14 = 0$ $ x^2 - 5x - 14 = 0$ Factor the expression: $ (x + 2)(x - 7) = 0$ Therefore $x = -2$ or $x = 7$ The original expression is defined at $x = -2$ and $x = 7$, so there are no extraneous solutions.